Let us say you decompose the Dirac field like the Klein-Gordon field, i.e.
$$\Psi(x)=\int[dk]\left(u(k)e^{-ikx}+v(k)e^{ikx}\right),$$
with the Lorentz-invariant integration measure $[dk]:=d^4k/(2\pi)^4\times2\pi\delta(k_0-\omega(\mathbf{k}))$, and define your positive- and negative-frequency solutions that way as $u,v$ respectively.
That means, if you interpret the solution $u$ as a Dirac fermion travelling in the direction $\mathbf{k}$ with positive frequency $k_0$, then $v$ is a Dirac fermion travelling into the direction $-\mathbf{k}$ with negative frequency $-k_0$. That is equivalent to saying that its Dirac-conjugate in $\int[dk]\overline{v}(k)e^{-ikx}$ is a Dirac-antifermion travelling in the direction $\mathbf{k}$ with positive frequency $k_0$.
Thus, while $u(k)$ is your Dirac-fermion, $\overline{v}(k)$ is your Dirac-antifermion. That interpretation makes sense since if you look at $e^+e^-$ annihilation, the incoming positron is described by the Dirac-conjugate $\overline{v}(k)$, and not $v(k)$.
So far so good, now we can look how the chirality projectors act on the Dirac-spinors. We define $u_{L/R}:=\frac{1}{2}(1\mp\gamma_5)u$ for the Dirac-fermion. Since $\overline{v}$ is the Dirac-antifermion, we should analogously define
$$\overline{v}_{L/R}:=\overline{v}\frac{1}{2}(1\mp\gamma_5)=v^\dagger\gamma^0\frac{1}{2}(1\mp\gamma_5)=v^\dagger\frac{1}{2}(1\pm\gamma_5)\gamma^0=\left(v_{R/L}\right)^\dagger\gamma^0=\overline{v_{R/L}}.$$
We see that the left-chiral Dirac-antifermion $\overline{v}_L=\overline{v_R}$ is a Dirac-conjugated right-chiral $v_R$. Note that the bars over the spinor have different lengths.
Now we do the last step and ditch the mass. Both $u$ and $v$ now fulfil the same Dirac equation $k_\mu\gamma^\mu u=0$, $k_\mu\gamma^\mu v=0$. The Dirac-spinors decompose into two decoupled Weyl-spinors $\hat{u}_{L/R}$, such that in the chiral representation $u_{L/R}=(\hat{u}_L,\hat{u}_R)^T$. These Weyl-spinors fulfil the Weyl equations
$$0=k_\mu\overline{\sigma}^\mu \hat{u}_L=k_0\hat{u}_L+k_i\sigma^i\hat{u}_L\leftrightarrow\frac{k_i\sigma^i}{|\mathbf{k}|}\hat{u}_L=-\hat{u}_L,$$
$$0=k_\mu\sigma^\mu \hat{u}_R=k_0\hat{u}_R-k_i\sigma^i\hat{u}_R\leftrightarrow\frac{k_i\sigma^i}{|\mathbf{k}|}\hat{u}_R=\hat{u}_R.$$
Here, we have used that $k_0=|\mathbf{k}|$ in the massless case to show that the chiral eigenstates are also helicity eigenstates for $m=0$: The left-chiral state has negative helicity, the right-chiral has positive helicity.
Since both $u$ and $v$ fulfill the same Dirac equation, the same two equations hold for the Weyl-spinors $\hat{v}_L,\hat{v}_R$. But as we should carefully remember, $v$ does not describe the antifermion - its Dirac-conjugated $\overline{v}$ does! In the chiral representation we have $\overline{v}=(\hat{v}^\dagger_L,\hat{v}^\dagger_R)$ and thus we have the Weyl equations
$$0=\hat{v}^\dagger_Rk_\mu\overline{\sigma}^\mu\leftrightarrow \hat{v}^\dagger_R\frac{k_i\sigma^i}{|\mathbf{k}|}=-\hat{v}^\dagger_R,$$
$$0=\hat{v}^\dagger_Lk_\mu\sigma^\mu\leftrightarrow \hat{v}^\dagger_L\frac{k_i\sigma^i}{|\mathbf{k}|}=\hat{v}^\dagger_L.$$
As we manifestly see, the left-chiral antifermion $\hat{v}^\dagger_L$ has positive helicity, unlike the left-chiral fermion $\hat{u}_L$. And the converse is true for the opposite chirality.
I must admit, I had to ponder a lot about this. The way I see it is that it is a typical case of 'be mindful about the maths in QFT and how you interpret it'. Indeed, the idea is that one must be careful that both the fermion and the antifermion field should propagate into the same direction. If you interpret the exponential $e^{-i(\omega t-\mathbf{kx})}$ as a planar wave moving in direction $\mathbf{k}$, then $u$ is your fermion field and $\overline{v}$ is your antifermion field travelling in the same direction. From then on it is clear that $\overline{v}$ will dictate the antifermion's properties, not $v$. And the same then extends to the massless limit where the Dirac-spinors $u,\overline{v}$ are replaced by the Weyl-spinors $\hat{u}_{L/R},\hat{v}^\dagger_{L/R}$.
